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How do you find the hypotenuse of a right triangle when only a side length and angle is given?
Dependent on what side you are given you would use Sin(Θ) = Opposite/Hypotenuse just rearrange the formula to Hypotenuse = Opposite/Sin(Θ). Or if you are given the adjacent side use Cosine(Θ)=Adjacent/Hypotenuse, then: Hypotenuse = Adjacent/Cosine(Θ)
What are the Names of all trigonometry functions?
The basic functions are sine, cosine, tangent, cosecant, secant and cotangent. In addition, there are their inverses, whose full names use the prefix "arc" [arcsine, arc cosine, etc] but are more often written as sin-1, cos-1 and so on.
How do you calculate sin20?
Sin(20) = You need either a scientific calculator or Castles four figure Tables. Using a scientific calculator Sin ( 20 ) = 0.342020143.... Sin(20) ~ 0.3420 (4d.p.).
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
Find the length of the side opposite angle a?
It depends on what else you know. If it is a non-right triangle, and you only know angle a, it is impossible to fing side A (the side opposite an angle usually has the same letter, but capitalized). If you know the other two sides, then I would use the law of cosines: For a triangle with sides A B C A = √(B2+C2-(2*B*C*(cos (a)))) If you know another angle and one side, I would use the law sines: A/(sin a) = B/(sin b) therefore, A = (sin a) * B/(sin b) If it is a right triangle, and you know another side, than your job is even easier: If you know the hypotenuse (side C), than: A = C *(sin a) If you know the adjacent side (side B), than: A = B * (tan a)
